# fourier analysis

Discussion in 'Math' started by u-will-neva-no, May 17, 2011.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hello,

Ive been staring at this maths problem for a LONG time now and having difficulties with the following question. Im going to attach the original question just in case I have missed anything important out.*note* I am referring to question a ii, the previous question has been solved and included just in case it helps for this one (not that I think it does...).

I have my teachers solution so im trying to piece a few things together. What i am trying to get my head around is how;

∫g(t)exp(jΩt) exp(-jwt)dt = G(ω-Ω) (The limits are ∞ and -∞ for the integral)

If i can understand this, then I will be able to understand the problem! Thanks

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2. ### fila Member

Feb 14, 2011
64
5
Join the two exp functions and you get exp[jt(Ω-ω)] or we can get the minus sign out so we have exp[-jt(ω-Ω)].
But ω = 2∏f and I presume Ω = 2∏F, where F is some defined value (frequency).
So exp[-j2∏t(f-F)] and if we substitute f - F with s than we have exp(-j2∏st). When you multiply this with the function g(t) and put that in the integral from -∞ to ∞ you will see a form that is a definition of the Fourier transform. You get G(s) or G(ω-Ω). I think.

Last edited: May 17, 2011
3. ### fila Member

Feb 14, 2011
64
5
I messed up a little. Maybe you can define s to be s = ω-Ω, but I am just learning Fourier and we always write frequencies f and not ω. But 2*∏ shouldn't make the difference. I hope you get the idea.

u-will-neva-no likes this.
4. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Im still slightly confused. Using what you said, I have∫ g(t)exp(-jt(ω-Ω)). When you say put that into the fourier transform, what equation is that referring to? I get you dont fully understand this like me, so I will generalise this to other members on this forum. Thanks for you help so far fila